Counting

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Problems about counting and probability.

Which of the following situations are the same type of counting situation?

The number of ways to see two threes on four rolls of a 6-sided die.. The number of ways to flip a coin four times and see two heads. The number of ways to elect a President and Vice President from a group of four people. The number of ways to choose two students from a class of four. The number of ways to choose two different scoops of ice cream from a shop offering four flavors. The number of ways to choose two scoops of ice cream which are the same flavor from a shop offering four flavors.

Use the Binomial Theorem to expand $(a+b)^4$ . How many terms are in the expansion? $\answer {5}$

Correct! Enter the terms in increasing powers of $b$ .

$\answer {a^4}+\answer {4a^3b}+\answer {6a^2b^2}+\answer {4ab^3}+\answer {b^4}$ .

Use the Binomial Theorem to expand $(3+x)^6$ . How many terms are in the expansion? $\answer {7}$

Correct! Enter the terms in increasing powers of $x$ .

$\answer {729}+\answer {1458x}+\answer {1215x^2}+\answer {540x^3}+\answer {135x^4}+\answer {18x^5}+\answer {x^6}$

Ignoring coefficients in the expansion of $(x+y)^9$ , which of the following terms will appear?

$x^2y^7$ $x^5y^3$ $x^4y^5$ $x^9y$ $y^9$ $x^6$ $xy^8$

Correct! All of the terms will be of degree $\answer {9}$ .

Furthermore, the coefficient of $x^2y^7$ will be $\answer {9\cdot 8/2}$ , and the coefficient of $x^6y^3$ will be $\answer {9\cdot 8\cdot 7/(2\cdot 3)}$ .

Explain why ${n \choose k} = {n \choose n-k}$ .

Using the context of the stop lights, $n \choose k$ represents $k$ green lights out of a total of $n$ lights. Remembering that $k$ green lights also means $n-k$ red lights, we could simply exchange the role of red lights and green lights, we see the result. Remember that you should be able to use two contexts to explain this pattern!

You flip a coin $5$ times. How many different ways are there to flip two heads? $\answer [given]{10}$

You flip a coin $5$ times. How many different ways are there to flip at least two heads? $\answer [given]{26}$

You flip a coin $5$ times. How many different outcomes are there? $\answer [given]{2^5}$

You flip a coin $5$ times. What is the probability that you flip exactly two heads? $\answer [given]{\frac {10}{32}}$

You flip a coin $5$ times. What is the probability that you flip at least two heads? $\answer [given]{\frac {26}{32}}$

You flip a coin $5$ times. What is the probability that your result was HHTTT? $\answer [given]{\frac {1}{32}}$

You flip a coin $5$ times, and you get two heads. What is the probability that your result was HHTTT? $\answer [given]{\frac {1}{10}}$

In your own words, summarize the similarities and differences between the previous problems.

The problems are mostly about “two-head” situations when flipping 5 coins. The differences are about whether we have exactly two heads versus at least two heads, or whether the heads occur for specific coins versus anywhere among the coins.

A certain passcode is made by choosing two digits in $0$ to $9$ followed by three shapes (square, triangle, circle, or star). How many such passcodes can be made? $\answer [given]{6400}$

A certain passcode is made by choosing two digits in $0$ to $9$ followed by three shapes (square, triangle, circle, or star). How many such passcodes can be made if you cannot choose the same number or same shape more than once? $\answer [given]{2160}$

A certain passcode is made by choosing five total symbols from the digits in $0$ to $9$ and the shapes in the collection (square, triangle, circle, or star). How many such passcodes can be made if you cannot choose the same number or same shape more than once, but you can choose the numbers and shapes in any order? $\answer [given]{240240}$

In your own words, summarize the similarities and differences between the previous problems.